Tehnička termodinamika - Kemijsko-tehnološki fakultet

_________________________________________________Drugi glavni zakon termodinamike
Rješenje
a)
R s
2
2m 3 n = = 0. 089 kmol
22.414
R s ∑ω
⋅ R
= n
i=
1
R s = ω CO ⋅ R + ω
i
2 CO2
N2
N2
i
⋅ R
8.314 8.314 0.270 kJ kg
−1 K
−1
= 0.25 ⋅ + 0.75 ⋅ =
44 28
M
c
s
R
= =
R
s
8.314
0.27
c p ∑ ⋅ c
= n
, s ωi
i=
1
= 30.79 kg kmol
p,
i
= ω ⋅ c + ω
⋅ c
−1
p, s CO2 p, CO2
N2
p N2
36.08 29.31
c p, s = 0.25 ⋅ + 0.75 ⋅ = 0.99 kJ kg
44 28
−1 K
−1
1 1
C p, s = c p,
s ⋅ M s = 0.99 ⋅ 30.79 = 30.48 kJ kmol
− K
−
−1
−1
cv, s = c p,
s − Rs
= 0.99 − 0.27 = 0.72 kJ kg K
−1
−1
Cv, s = cv,
s ⋅ M s = 0.72 ⋅ 30.79 = 22.17 kJ kmol K
W
max.
= U
1
−U
2
− T
0
⋅
( s − s ) + p ⋅ ( V −V
)
1
2
( T − T ) = 0.089 ⋅ 22.17 ⋅ ( 300 − 288) 24
U1 −U
2 = ns
⋅ Cv,
s ⋅ 1 2
=
⎛
T
p
T0 ⋅
ln
⎝
( ) ⎜
1
1
s1
− s2
= T0
⋅ ns
⋅ C p,
s ⋅ ln − ns
⋅ R ⋅
T
⎟ 2
p2
⎠
⎛
300
210 ⎞
T0 ⋅ 1 2 ⎜
⎟ −
⎝
288
110 ⎠
( s − s ) = 288 ⋅ 0.089 ⋅ 30.48 ⋅ ln − 0.089 ⋅ 8.314 ⋅ ln = 106
0
1
2
⎞
kJ
kJ
191